Given $a,b,c$ are the sides of a triangle. Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$

Given $a,b,c$ are the sides of a triangle. Prove that $\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}<2$.

My attempt:
I could solve it by using the semiperimeter concept. I tried to transform this equation since it is a homogeneous equation as $f(a,b,c)=f(ta,tb,tc)$. I considered $a+b+c=1$ and thus the inequality reduces to

$$\dfrac{a}{1-a}+\dfrac{b}{1-b}+\dfrac{c}{1-c}<2$$.

This is getting quite difficult to prove, as I don’t have any clue to approach. I want to solve this only by homogeneous equation transformation, so any other transformation is welcome, but not any other idea(I don’t mean to be rude but that’s my necessity). Please help. Thank you.

Solutions Collecting From Web of "Given $a,b,c$ are the sides of a triangle. Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$"

As these are sides of a triangle, let $a=x+y, b= y+z, c=z+x$, and using homogeneity, set $x+y+z=1$. The inequality is then to show:
$$\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z} < 2$$

Note that $x \in (0, 1) \implies 1+x > 1 \implies \dfrac{1-x}{1+x}< 1-x$. Sum that across $x, y, z$ to get the above inequality.

Alternatively, without much multiplying: by triangle inequality you get $\frac{a}{b+c} < 1$ and hence
$$\frac{a}{b+c} < \frac{a + a}{a + b + c} = \frac{2a}{a + b + c}$$
Adding 3 similar inequalities (with respectively $a, b, c$ in the numerators on the left hand side fractions) will prove the thesis.

it is equivalent to
$a^2(b+c-a)+b^2(a+c-b)+c^2(a+b-c)+abc>0$